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4.9t^2-3t-9.5=0
a = 4.9; b = -3; c = -9.5;
Δ = b2-4ac
Δ = -32-4·4.9·(-9.5)
Δ = 195.2
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{195.2}}{2*4.9}=\frac{3-\sqrt{195.2}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{195.2}}{2*4.9}=\frac{3+\sqrt{195.2}}{9.8} $
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